Question

Помогите с пятым заданием

Answer 1

Задание 1

а) $240000 = 24 \cdot 10^4$

б) $0,00043 = 43 \cdot 10^{-5}$

в) $214 = 0,214 \cdot 10^3$

Задание 2

а) $(2,5 \cdot 10^{12}) \cdot (3,2 \cdot 10^{-5}) = 8 \cdot 10^{12+(-5)} = 8\cdot 10^{12-5} = 8\cdot 10^7$

б) $(2,5 \cdot 10^{12}) : (3,2 \cdot 10^{-5}) = 0,78125 \cdot 10^{12-(-5)} = 0,78125 \cdot 10^{12+5} = 0,78125 \cdot 10^{17}$

Задание 3

$(b^{-2} - a^{-2})\cdot (\frac{a+b}{ab})^{-1} = (\frac{1}{b^2} - \frac{1}{a^2}) \cdot \frac{ab}{a+b} = \frac{a^2-b^2}{a^2b^2} \cdot \frac{ab}{a+b} = \frac{ab(a-b)(a+b)}{a^2b^2(a+b)} = \frac{a-b}{ab}$

Задание 4

$\frac{x^{-2}+y^{-2}}{(x+y)^2} + \frac{2x^{-1}+2y^{-1}}{(x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{\frac{1}{x^2} + \frac{1}{y^2}}{(x+y)^2} + \frac{\frac{2}{x} + \frac{2}{y}}{(x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{(\frac{1}{x^2} + \frac{1}{y^2})(x+y)}{(x+y)^3} + \frac{\frac{2}{x} + \frac{2}{y}}{(x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{(\frac{x+y}{x^2} + \frac{x+y}{y^2})}{(x+y)^3} + \frac{\frac{2}{x} + \frac{2}{y}}{(x+y)^3} = \frac{1}{x^2y^2}$

$\frac{\frac{y^2(x+y)+x^2(x+y)+2xy^2+2x^2y}{x^2y^2}}{(x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{\frac{xy^2+y^3+x^3+x^2y+2xy^2+2x^2y}{x^2y^2}}{(x+y)^3} = \frac{1}{x^2y^2} \\\\ \frac{\frac{x^3 + 3x^2 + 3xy^2 + y^3}{x^2y^2}}{(x+y)^3} = \frac{1}{x^2y^2} \\\\\frac{\frac{(x+y)^3}{x^2y^2}}{(x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{(x+y)^3}{x^2y^2} : (x+y)^3 = \frac{1}{x^2y^2}\\\\\frac{(x+y)^3\cdot 1}{x^2y^2\cdot (x+y)^3} = \frac{1}{x^2y^2}\\\\\frac{1}{x^2y^2} = \frac{1}{x^2y^2}$